LEC ANA2 15

More on path connectedness

Recall Exm 337.6.

Example 339.1(Example).

Let

$$E=\left\{\left(x,\sin \left(\frac{1}{x}\right)\right):x\in (0,1]\right\}.$$

$\overline{E}=E\cup \{0\}\times [-1,1]$. Clearly, $E$ is connected (in fact, path connected), and hence $\overline{E}$ is connected. However, $\overline{E}$ is not path connected.

Indeed, suppose there is a path $\gamma$ connecting $(0,0)$ and $(1,\sin (1))$. Let $K={\gamma }^{-1}(\{0\}\times [-1,1])\subseteq [0,1]$. Since $K$ is compact, we have $t_0:=\sup K\in K$; note that $t_0\ne 1$. For all $t>t_0$, $\gamma (t)\in E$. Thus, ${\pi }_x(\gamma (t_0))=0$, whereas ${\pi }_x(\gamma (t_0+1/k))>0$ for all $k>0$.

Fix $\theta \in [0,2\pi )$ such that $\sin \theta \ne {\pi }_y(\gamma (t_0))$. Let $\{t_i\}\subseteq (t_0,1]$ be such that $\{t_i\}\to t_0$. For each $t_i$, there exists $n_i\in \mathbb{N}$ such that

$${\pi }_x(\gamma (t_0))=0<\frac{1}{2\pi n_i+\theta }<{\pi }_x(\gamma (t_i)).$$

By the intermediate value theorem, there exists $0<r_i<t_i$ such that ${\pi }_x(\gamma (r_i))=\frac{1}{2\pi n_i+\theta }$. It follows that $\{{\pi }_y(\gamma (r_i))\}\to \sin \theta$, a contradiction.

Proposition 339.2(Proposition).

The image of a path connected set under a continuous map is path connected.

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Lemma 339.3(Lemma).

Let $p$ be a complex polynomial in $n$ complex variables. Let $Z(p)$ be the zero set of $p$. ${\mathbb{C}}^n\setminus Z(p)$ is path connected.

Proof.

Let $\mathbf{z},\mathbf{w}\in {\mathbb{C}}^n\setminus Z(p)$. Define $\gamma :\mathbb{C}\to {\mathbb{C}}^n$ by

$$\gamma (z)=z\mathbf{z}+(1-z)\mathbf{w}.$$

$p(\gamma (z))$ is a polynomial in one variable, so $Z(p\circ \gamma )\subseteq \mathbb{C}$ is a finite set. Therefore, $\mathbb{C}\setminus Z(p\circ \gamma )$ is path connected. By Prp 2, $\gamma (\mathbb{C}\setminus Z(p\circ \gamma ))\subseteq {\mathbb{C}}^n\setminus Z(p)$ is path connected. Observe that $\mathbf{z},\mathbf{w}\in \gamma (\mathbb{C}\setminus Z(p\circ \gamma ))$.□

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Example 339.4(Example).

It is immediate from Lem 3 that $G{L}_n(\mathbb{C})$ is path connected.

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Cantor set

Definition 339.5(Definition).

Define ${\mathcal{C}}_0=[0,1]$. Express ${\mathcal{C}}_n$, $n⩾0$, as

$${\mathcal{C}}_n=\coprod _{i\in I_n}{J}_i^n,$$

where $|I_n|=2^n$ and

$$J_i^n=\left[\frac{i}{3^n},\frac{i+1}{3^n}\right].$$

Define

$${\mathcal{C}}_{n+1}:=\coprod _{i\in I_n}\left[\frac{3i}{3^{n+1}},\frac{3i+1}{3^{n+1}}\right]\cup \left[\frac{3i+2}{3^{n+1}},\frac{3i+3}{3^{n+1}}\right].$$

Define the Cantor set $\mathcal{C}$ to be

$$\mathcal{C}:=\bigcap _{i=0}^{\infty }{\mathcal{C}}_i.$$

$\mathcal{C}$ is clearly non-empty (it contains the endpoints of all $J_i^n$) and compact.

Proposition 339.6(Proposition).

$\mathcal{C}$ is perfect.

Proof.

We have to show that every $x\in \mathcal{C}$ is a limit point of $\mathcal{C}$. This is clear, since for $ϵ>0$, we can choose $n$ such that $3^n<ϵ$, forcing an endpoint of some $J_i^n$ to lie in $(x-ϵ,x+ϵ)$.□

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It follows from Exr 291.4 and Prp 6 that $\mathcal{C}$ is uncountable.

Proposition 339.7(Proposition).

$\mathcal{C}$ is homeomorphic to $\{0,2{\}}^{\mathbb{N}}$ with the product topology.

Proof.

Observe that each $x\in \mathcal{C}$ has a unique expression of the form

$$\begin{array}{rlr}x=\sum _{i=1}^{\infty }\frac{a_i}{3^i}& & (a_k\in \{0,2\}),\end{array}$$

since $0.0\overline{2}=0.1$. This allows us to define a bijection $\phi :\mathcal{C}\to \{0,2{\}}^{\mathbb{N}}$. By Thm 15.5, a sequence in $\{0,2{\}}^{\mathbb{N}}$ converges iff it converges coordinate-wise. Thus, to show $\phi$ is continuous, it suffices to show ${\pi }_i\circ \phi$ is continuous for each $i\in \mathbb{N}$, which is immediate since these are locally constant maps. Since $\mathcal{C}$ is compact, it follows from Thm 143.2 that $\phi$ is a homeomorphism.□

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Prp 7 can be seen an another proof of the uncountability of $\mathcal{C}$.

Proposition 339.8(Proposition).

${\mathcal{C}}^{\mathbb{N}}$ is homeomorphic to $\mathcal{C}$.

Proof.

We have to show $\{0,2{\}}^{\mathbb{N}}\cong (\{0,2{\}}^{\mathbb{N}}{)}^{\mathbb{N}}$. Let $\phi$ be the diagonal bijection $\{0,2{\}}^{\mathbb{N}}\to (\{0,2{\}}^{\mathbb{N}}{)}^{\mathbb{N}}$:

$${\mathbf{x}}_n=(x_n^1,x_n^2,\dots )\mapsto \left(\begin{array}{ccccccccccc}(& x_n^1& ,& x_n^3& ,& x_n^6& ,& x_n^{10}& ,& \dots & ),\\ (& x_n^2& ,& x_n^5& ,& x_n^9& ,& \dots & & & ),\\ (& x_n^4& ,& x_n^8& ,& \dots & & & & & ),\\ (& x_n^7& ,& \dots & & & & & & & ),\\ & & & & & ⋮& & & & & \end{array}\right)=:\left(\begin{array}{c}{\mathbf{z}}_{1,n}\\ {\mathbf{z}}_{2,n}\\ {\mathbf{z}}_{3,n}\\ {\mathbf{z}}_{4,n}\\ ⋮\end{array}\right)$$

If a sequence $\{{\mathbf{x}}_n{\}}_{n=1}^{\infty }$ converges in $\{0,2{\}}^{\mathbb{N}}$, it must converge coordinate-wise, that is, each $\{x_n^i\}$ converges as $n\to \infty$. In particular, $x_n^1,x_n^3,x_n^6,\dots$ converge coordinate-wise, so ${\mathbf{z}}_{1,n}$ converges. Similarly, ${\mathbf{z}}_{k,n}$ converge as $n\to \infty$ for all $k$. Therefore, the sequence $\{({\mathbf{z}}_{1,n},{\mathbf{z}}_{2,n},\dots ){\}}_{n=1}^{\infty }$ converges “coordinate-wise”, and hence converges in $(\{0,2{\}}^{\mathbb{N}}{)}^{\mathbb{N}}$. Thus, $\phi$ is continuous. That $\phi$ is a homeomorphism follows form the compactness of $\{0,2{\}}^{\mathbb{N}}$ and Thm 143.2.□

Proposition 339.9(Proposition).

There exists continuous and surjective $f:\mathcal{C}\to [0,1]$.

Proof.

Define$f$ by

$$\begin{array}{r}f\left(\sum _{i=1}^{\infty }\frac{a_i}{3^i}\right)=\sum _{i=1}^{\infty }\frac{a_i}{2^{i+1}}.\end{array}$$

Let $x\in \mathcal{C}$ and $f(x)=y$. Let $ϵ>0$. Choose $n$ such that $1/2^n<ϵ$. Let $2\delta =1/3^n$. The ternary representations of all elements in $(x-\delta ,x+\delta )\cap \mathcal{C}$ will have their first $n$ digits fixed. If $x^{\prime }$ is such an element, then

$$|f(x)-f(x^{\prime })|⩽\sum _{i=n+1}^{\infty }\frac{1}{2^i}=\frac{1}{2^n}<ϵ.$$

Thus, $f$ is continuous.□

Note that $f$ is not injective, since $f(0.0\overline{2})=f(0.2)$.

Fact 339.10(Fact).

Any compact metric space is a continuous image of $\mathcal{C}$¹.

Proposition 339.11(Proposition).

$\mathcal{C}$ is nowhere dense in $[0,1]$.

Definition 339.12(Definition).

$S\subseteq X$ is totally disconnected if for $x,y\in S$, $x\ne y$, there exist open $U_x\ni x$ and $U_y\ni y$ such that $U_x\cap U_y=\varnothing$ and $U_x\bigsqcup U_y\supseteq S$.

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Proposition 339.13(Proposition).

$\mathcal{C}$ is totally disconnected.

Proof.

Suppose $x,y\in \mathcal{C}$, $x\ne y$. There must exist $J_i^n$ such that $x\in J_i^n$ and $y\notin J_i^n$². Now,

$$J_i^n\cap \mathcal{C}=\left(\frac{i}{3^n}-\frac{1}{3^{n+1}},\frac{i+1}{3^n}+\frac{1}{3^{n+1}}\right)\cap \mathcal{C}.$$

It follows³ that $J_i^n\cap \mathcal{C}$ is clopen in $\mathcal{C}$. $U_x:=J_i^n\cap \mathcal{C}$, $U_y:=\mathcal{C}\setminus (J_i^n\cap \mathcal{C})$ satisfy Def 12.□

Footnotes

1. Idea: Cover with $2^k$ many balls, where $k$ is fixed for each level? ↩

2. this follows from the fact that $\mathcal{C}$ is compact and hence has the Cantor intersection property: Cantor intersection theorem. ↩

3. Subspace topology ↩